How do you find the integral of #cos(x)^2*sin(x)^2#?

Question

Emilia Aguilar · Accepted Answer

#x/8-1/32sin(4x)+C#

This can be written as:

#I=intcos^2(x)sin^2(x)dx#

We can rewrite this using the identity #sin(2x)=2sin(x)cos(x)#. Thus, #sin(x)cos(x)=sin(2x)/2#. Square both sides to see that #cos^2(x)sin^2(x)=sin^2(2x)/4#. Thus:

#I=1/4intsin^2(2x)dx#

Let #u=2x# so that #du=2dx#.

#I=1/8intsin^2(2x)*(2dx)=1/8intsin^2(u)du#

The way to integrate this is to use another double-angle formula.

#cos(2u)=1-2sin^2(u)" "=>" "sin^2(u)=1/2(1-cos(2u))#

Thus:

#I=1/16int(1-cos(2u))du#

#I=1/16intdu-1/16intcos(2u)du#

The first integral is the most basic integral and the second can be solved through inspection or substitution, where #v=2u#.

#I=1/16u-1/32sin(2u)+C#

Since #u=2x#:

#I=x/8-1/32sin(4x)+C#

Samantha Adkison · Answer

undefined To find the integral of cos(x)^2*sin(x)^2, you can use the double angle identity for cosine:

cos(2x) = 2*cos(x)^2 - 1

Now, rewrite cos(x)^2 as (1 + cos(2x))/2:

(1 + cos(2x))/2 * sin(x)^2

Use the identity sin(2x) = 2*sin(x)*cos(x):

(1 + cos(2x))/2 * (sin(2x)/2)

Now integrate with respect to x:

∫ [(1 + cos(2x))/2 * (sin(2x)/2)] dx

= (1/4) ∫ [(sin(2x) + sin(2x)*cos(2x))] dx

= (1/4) [(-1/2)*cos(2x) + (1/4)*sin(2x)^2] + C

Where C is the constant of integration.